How do you define $e_p(0)$? Presumably as $\infty$? Also, the fact that $\mathbb{Z}_{p^k}[X]$ is not a UFD need not be an obstacle; if the polynomial is irreducible modulo $p^k$, then it is necessarily irreducible modulo $p^k$. The concept of irreducibility still makes sense in $\mathbb{Z}_{p^k}[X]$.
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Arturo MagidinDec 21 '10 at 16:22

@Arturo: Oops, my bad! You're right, I should've added that. Fortunately this omission isn't too confusing. Also, I was trying to argue somewhere along these lines: reduce f mod $p^k$ (k as big as possible) to obtain a polynomial of the form $cx^n$. Now, had $Z_{p^k}[X]$ been a UFD, the only possible factorization would be into polynomials $dx^s$ and $ex^{k-s}$...
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swanDec 22 '10 at 12:38

2 Answers
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Why would irreducibility of $\frac{1}{p^{n-1}}x^nf(\frac{p}{x})$ imply irreducibility of f? Also I have to admit I don't really know how to interpret $f(\frac{p}{x})$.
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swanDec 22 '10 at 12:47

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@swan: $x^nf(\frac{1}{x})$ is the polynomial you get by 'reversing' the powers. If you had $a_nx^n + a_{n-1}x^{n-1}+\cdots+a_0$ (with $a_n\neq 0$, then $x^nf(\frac{1}{x}) = a_n + a_{n-1}x + \cdots + a_1x^{n-1} + a_0x^n$. Now see what happens if you do $x^nf(\frac{p}{x})$ (write out the powers of $p$ in the $a_i$ explicitly, that will help), and then see if you can factor out $p^{n-1}$. Factorizations of $f(x)$ correspond to factorizations of $x^nf(\frac{1}{x})$ in the "obvious way" (just 'reverse the powers' in the factors).
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Arturo MagidinDec 23 '10 at 6:45

OK, I get in now. Thanks a lot! I'm accepting this one since I understand it better.
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swanDec 29 '10 at 12:08

One way to prove the irreducibility seems to be to use the Newton Polygon.
The condition on the coefficients of $f$ means that the Newton Polygon has a side of slope $\dfrac{1}{n}-1$ and hence that $f$ has a root $\alpha$ in some algebraic closure $F$ of $\mathbf{Q}_p$ having valuation $1 - \dfrac{1}{n}$ (there is a unique way to prolong $e_p$ to a valuation of $F$).

But then the extension $\mathbf{Q}_p \subset \mathbf{Q}_p(\alpha)$ is totally ramified
of degree $n$ and $f$ must be irreducible over $\mathbf{Q}_p$ hence a fortiori irreducible over $\mathbf{Q}$.

Thanks for that! Gotta learn about the Newton Polygons. I wonder if there's a more elementary way of proving validity of the criterion though. I mean, Newton Polygons are not something I could've came up with myself.
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swanDec 22 '10 at 12:46

@swan: Right. Well, Zarrax's solution is more elementary. But the Newton polygon seems to work pretty "naturally" for this problem. There is a treatment of the NP in the 1st chapter of Dwork, Gerotto and Sullivan "An Intro to G-Functions".
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George McNinchDec 24 '10 at 0:09

Great! I'll definitely take a look. Thank you once more!
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swanDec 29 '10 at 12:09